Question Number 218345 by Spillover last updated on 07/Apr/25
Answered by som(math1967) last updated on 07/Apr/25
$${let}\:{AF}={AE}={x} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{3}×\left(\mathrm{6}+{x}\right)+\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{3}×\left(\mathrm{8}+{x}\right)+\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{3}×\mathrm{14}=\mathrm{63} \\ $$$$\Rightarrow\frac{\mathrm{3}}{\mathrm{2}}×\left(\mathrm{28}+\mathrm{2}{x}\right)=\mathrm{63} \\ $$$$\mathrm{28}+\mathrm{2}{x}=\mathrm{42} \\ $$$$\:\therefore{x}=\mathrm{7} \\ $$$$\:{AB}=\mathrm{6}+\mathrm{7}=\mathrm{13}{cm} \\ $$
Commented by Spillover last updated on 08/Apr/25
$${thank}\:{you} \\ $$